Step-by-step explanation:
Hey, there!!!
Let me simply explain you about it.
We generally use the distance formula to get the points.
let the point R be (x,y)
As it an equilateral triangle it must have equal distance.
now,
let's find the distance of PQ,
we have, distance formulae is;
[tex]pq = \sqrt{( {x2 - x1)}^{2} + ( {y2 - y1)}^{2} } [/tex]
[tex]or \: \sqrt{( {5 + 3)}^{2} + ( {2 - 2)}^{2} } [/tex]
By simplifying it we get,
[tex] 8[/tex]
Now,
again finding the distance between PR,
[tex] pr = \sqrt{( {x2 - x1}^{2} + ( {y2 - y1)}^{2} } [/tex]
or,
[tex] \sqrt{( {x + 3)}^{2} + ( {y - 2)}^{2} } [/tex]
By simplifying it we get,
[tex] = \sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 } [/tex]
now, finding the distance of QR,
[tex]qr = \sqrt{( {x - 5)}^{2} + ( {y - 2)}^{2} } [/tex]
or, by simplification we get,
[tex] \sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 } [/tex]
now, equating PR and QR,
[tex] \sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13} = \sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 } [/tex]
we cancelled the root ,
[tex] {x}^{2} + {y}^{2} + 6x - 4y + 13 = {x}^{2} + {y}^{2} -10x - 4y + 29[/tex]
or, cancelling all like terms, we get,
6x+13= -10x+29
16x=16
x=16/16
Therefore, x= 1.
now,
equating, PR and PQ,
[tex] \sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 } = 8} [/tex]
cancel the roots,
[tex] {x}^{2} + {y}^{2} + 6x - 4y + 13 = 8[/tex]
now,
(1)^2+ y^2+6×1-4y+13=8
or, 1+y^2+6-4y+13=8
y^2-4y+13+6+1=8
or, y(y-4)+20=8
or, y(y-4)= -12
either, or,
y= -12 y=8
Therefore, y= (8,-12)
by rounding off both values, we get,
x= 1
y=(8,-12)
So, i think it's (1,8) is your answer..
Hope it helps...
